I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$
$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{2 \sqrt[3]{x^2}}$$
But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)
UPDATE
I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)
$\endgroup$ 35 Answers
$\begingroup$Here is a hint: Use the identity $(a^3-b^3)=(a-b)\cdot(a^2+ab+b^2)$ with $a$, $b$ being suitable cube roots. Otherwise, the method is similar to the one you tried.
$\endgroup$ 2 $\begingroup$$$\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} $$ $$=\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} \cdot \frac{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}} $$
$$=\lim_{h \to 0} \frac{x+h-x}{h((x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3})}$$ $$=\lim_{h \to 0} \frac{1}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}$$ $$=\frac{1}{(x)^{2/3} + x^{1/3}(x)^{1/3} + (x)^{2/3}}$$ $$=\frac{1}{3x^{2/3}}$$ $$=\frac{x^{-2/3}}{3}$$ As obtained from the $Dx^{n} = n.x^{n-1}$
$\endgroup$ 1 $\begingroup$I understand that the point of this exercise is to apply the limit definition of the derivative to a function where the limit calculation is "tricky". But it's worth noting that if $F(x,y)=0$ identically (as in $y-\sqrt[3]{x}=0$ in this problem) then $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$.
So given that $x=y^3$, we have that $\frac{dx}{dy}=3y^2$ (either using the power rule or a simpler limit computation). That makes $\frac{dy}{dx}=\frac{1}{3y^2}=\frac{1}{3(\sqrt[3]{x})^2}=\frac{1}{3}x^{-\frac{2}{3}}$.
$\endgroup$ $\begingroup$You can use a similar 'trick' to find the derivative of $ y=\sqrt[n] x $. The limit will involve multiplying the numerator $(x+h)^{1/n} - x^{1/n} $ by an appropriate expression to get $ (x + h) - x $ .
$\endgroup$ 1 $\begingroup$Note that this works to find the derivative of $x^{1/n}$ where $n$ is a positive integer.
We use $a^n-b^n =(a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k} $.
$\begin{array}\\ \frac{(x+h)^{1/n}-x^{1/n}}{h} &=\frac{(x+h)^{1/n}-x^{1/n}}{h} \frac{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\ &=\frac{(x+h)-x}{h} \frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\ &=\frac{h}{h} \frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\ &=\frac{1}{\sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n}}\\ \end{array} $
As $h \to 0$,
$\begin{array}\\ \sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n} &\to \sum_{k=0}^{n-1} x^{k/n} x^{(n-1-k)/n}\\ &= \sum_{k=0}^{n-1} x^{(n-1)/n}\\ &= n x^{(n-1)/n}\\ &= n x^{1-1/n}\\ \end{array} $
Therefore $(x^{1/n})' =\frac1{n x^{1-1/n}} =\frac1{n} x^{1/n-1} $.
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